The `21 cm` radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

Options

A. `10^(-17)`
B. `1 J`
C. `7 xx 10^(-6)J`
D. `10^(-24)`J

Answer

To find the energy of the emitted 21 cm radio wave from hydrogen due to hyperfine interaction, we can use the formula for the energy of a photon, which is given by:

\[ E = h \nu \]

Where:
- \( E \) is the energy,
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \),
- \( \nu \) is the frequency of the wave.

We can also express frequency in terms of wavelength (\( \lambda \)) using the relationship:

\[ \nu = \frac{c}{\lambda} \]

Where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \),
- \( \lambda \) is the wavelength.

Given that the wavelength of the emitted wave is \( 21 \, \text{cm} \), we first convert this into meters:

\[ \lambda = 21 \, \text{cm} = 21 \times 10^{-2} \, \text{m} \]

Now we can substitute this value into the frequency equation:

\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{21 \times 10^{-2} \, \text{m}} \]

Calculating \( \nu \):

\[ \nu = \frac{3 \times 10^8}{0.21} \approx 1.42857 \times 10^{9} \, \text{Hz} \]

Now we can substitute the frequency back into the energy equation:

\[ E = h \nu = (6.626 \times 10^{-34} \, \text{J s}) \times (1.42857 \times 10^{9} \, \text{Hz}) \]

Calculating \( E \):

\[ E \approx 9.466 \times 10^{-25} \, \text{J} \]

Rounding this value gives us approximately:

\[ E \approx 1 \times 10^{-24} \, \text{J} \]

Thus, the energy of the emitted wave is nearly \( 1 \times 10^{-24} \, \text{J} \).

### Final Answer:
The energy of the emitted wave is nearly \( 1 \times 10^{-24} \, \text{J} \).